Problem: Subtract. $\dfrac{8}{3} - \dfrac{6}{4} = $
Answer: Before we can subtract our fractions, they need to have the same denominator. $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{4}$ $\frac{1}{4}$ $\frac{1}{4}$ $\frac{1}{4}$ $\frac{1}{4}$ $\frac{1}{4}$ $\frac{1}{4}$ $\frac{1}{4}$ $\frac{1}{4}$ $\frac{1}{4}$ $\frac{1}{4}$ $\frac{1}{4}$ $\dfrac{8}{3}$ $\dfrac{6}{4}$ $\dfrac{8}{3}-\dfrac{6}{4}$ Let's look at the multiples of each denominator and see which multiples they have in common. Denominator Multiples ${3}$ $3, {6}, 9, \underline{12}$ $4}$ $4, 8, \underline{12}, 16$ The least common denominator is ${12}$. Let's use multiplication to make each fraction have a denominator of $12$. ${\dfrac{8}{3}}=\dfrac{{8} \times 4}{{3} \times 4} = {\dfrac{32}{12}}$ $\dfrac{6}{4}}=\dfrac{6} \times 3}{4} \times 3} = {\dfrac18}12}}$ Now, we can subtract ${\dfrac{32}{12}} - \dfrac{18}{12}}$. $\dfrac{32}{12}$ $\dfrac{18}{12}$ $\dfrac{32}{12} - \dfrac{18}{12}$ $=\dfrac{{32}-18}}{12}$ $=\dfrac{14}{12}$ ${\dfrac{8}{3}} - \dfrac{6}{4}} = \dfrac{14}{12}$ We can also write $\dfrac{14}{12}$ as $\dfrac76$ or $1\dfrac16$.